About minimal geodesic of product of Riemannian space

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I'm struggling to prove the following :

let $M = L \times N$ be the product of two complete Riemannian manifolds $(L,g)$ and $(N,h)$, with the metric $ (g\times h)(v,w) = g_{p'}(v',w') + h_{p''}(v'',w'')$ Where $T_{p}M = T_{p'}L \times T_{p''}N$. Then, any geodesic on $M$ is minimal if and only if the two component geodesics on $L$ and $N$ are minimal.

Here is the source of my confusion :

If $\gamma$ is a minimal geodesic, I write $\gamma_{1}$ and $\gamma_{2}$ its projections on $L$ and $N$ respectively. then it minimizes the length $L(\gamma)=\int_{0}^{1}\sqrt{g_{\gamma_{1}(t)}(\dot{\gamma_{1}(t)},\dot{\gamma_{1}(t)}) + h_{\gamma_{2}(t)}(\dot{\gamma_{2}(t)},\dot{\gamma_{2}}(t))}dt$.

But for me, it is not clear how minimizing this quantity is related to minimizing both $\int_{0}^{1}\sqrt{g_{\gamma_{1}(t)}(\dot{\gamma_{1}(t)},\dot{\gamma_{1}(t)})}dt$ and $\int_{0}^{1}\sqrt{g_{\gamma_{2}(t)}(\dot{\gamma_{2}(t)},\dot{\gamma_{2}(t)})}dt$.

For instance, I tried to first prove some property like :

For every positive function $f,f',g$ if $\int_{0}^{1} \sqrt{f} > \int_{0}^{1} \sqrt{f'}$ then $ \int_{0}^{1} \sqrt{g+f} > \int_{0}^{1} \sqrt{g+f'}$

But I don't think this is true.

Please enlighten me !

Thanks

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Assuming you already are able to show that $\gamma_1$ and $\gamma_2$ are themselves (possibly non-minimizing) geodesics, notice that $g_{\gamma_1(t)}(\dot{\gamma}_1(t), \dot{\gamma}_1(t))$ is a constant, and similarly for $\gamma_2$. It follows that your integrand is a constant. This lets you deal easily with the otherwise annoying-looking square root!

From this you can show a generalized Pythagorean theorem, namely $$ \big(L(\gamma)\big)^2 = \big(L(\gamma_1)\big)^2 + \big( L(\gamma_2) \big)^2 $$ and the proof should be straightforward from there.